CBSE Class 10 Sample Paper 2024 Solution for Maths Standard

CENTRAL BOARD OF SECONDARY EDUCATION

SAMPLE PAPER
CLASS 10
2024

Solutions

Marking Scheme
Class X Session 2023-24
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
SECTION A
Section A consists of 20 questions of 1 mark each.
1. (b) xy2 1
2. (b) 1 zero and the zero is ‘3’ 1
3. 𝑎1 𝑏1 𝑐1 1
(b) = ≠ 𝑐2
𝑎2 𝑏2
4. (c) 2 distinct real roots 1
5. (c) 7 1
6. (a) 1:2 1
7. (d) infinitely many 1
8. 𝑎𝑐 1
(b)
𝑏+𝑐
9. (b) 100° 1
10. (d) 11 cm 1
11. √𝑏 2 −𝑎2 1
(c)
𝑏
12. (d) cos A 1
13. (d) 60° 1
14. (a) 2 units 1
15. (a) 10m 1
16. 4−𝜋 1
(b)
4
17. 22 1
(b)
46
18. (d) 150 1
19. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of 1
assertion (A)
20. (c) Assertion (A) is true but reason (R) is false. 1
SECTION B
Section B consists of 5 questions of 2 marks each.
21. Let us assume, to the contrary, that √2 is rational.
𝑎 ½
So, we can find integers a and b such that √2 = 𝑏 where a and b are coprime.
So, b √2 = a.
Squaring both sides,
we get 2b2 = a2. ½
Therefore, 2 divides a2 and so 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. ½
This means that 2 divides b2, and so 2 divides b
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1. ½
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.

1

22. ABCD is a parallelogram. ½
AB = DC = a
Point P divides AB in the ratio 2:3
2 3
AP = a , BP = a
5 5
point Q divides DC in the ratio 4:1. ½
4 1
DQ = a , CQ = a
5 5
∆ APO ~ ∆ CQO [AA similarity]
𝐴𝑃 𝑃𝑂 𝐴𝑂 ½
= =
𝐶𝑄 𝑄𝑂 𝐶𝑂
2 ½
AO a 2
CO
= 5
1 = 1 ⇒ OC = ½ OA
a
5
23.
PA = PB; CA = CE; DE = DB [Tangents to a circle] ½
Perimeter of ∆PCD = PC + CD + PD
= PC + CE + ED + PD
= PC + CA + BD + PD
= PA + PB 1
Perimeter of ∆PCD = PA + PA = 2PA = 2(10) = 20 ½
cm
24. ∵ tan(𝐴 + 𝐵) = √3 ∴ 𝐴 + 𝐵 = 600 …(1) ½
∵ tan(𝐴 − 𝐵) =
1
∴ 𝐴 − 𝐵 = 30 0
…(2) ½
√3 ½
Adding (1) & (2), we get 2A=900 ⟹ 𝐴 = 450 ½
Also (1) –(2), we get 2𝐵 = 300 ⟹ 𝐵 = 450
3
2 cosec230 + x sin260 – tan230 = 10
4
2 2
√3 3 1
⇒ 2(2)2 + x ( ) − 4 ( ) = 10 1
2 √3
3 3 1
⇒ 2(4) + x (4) − 4 (3) = 10 ½
3 1
⇒ 8 + x (4) − 4 = 10
⇒ 32 + x (3) −1 = 40 ½
⇒ 3x = 9 ⇒ x = 3
25. ∠𝐴 ∠𝐵 ∠𝐶
Total area removed = 360 π𝑟 2 + 360 π𝑟 2 + 360 π𝑟 2 ½
∠𝐴+ ∠𝐵+ ∠𝐶
= π𝑟 2
360
180
= 360 π𝑟 2 ½
180 22
= X X (14)2 ½ ½
360 7
= 308 cm2

The side of a square = Diameter of the semi-circle = a
Area of the unshaded region ½
= Area of a square of side ‘a’ + 4(Area of a semi-circle of diameter ‘a’)
The horizontal/vertical extent of the white region = 14-3-3 = 8 cm ½
Radius of the semi-circle + side of a square + Radius of the semi-circle = 8 cm

2

2 (radius of the semi-circle) + side of a square = 8 cm
2a = 8 cm ⇒ a = 4 cm ½
Area of the unshaded region
= Area of a square of side 4 cm + 4 (Area of a semi-circle of diameter 4 cm)
1
= (4)2 + 4 X 2 π(2)2 = 16 + 8π cm2 ½
SECTION C

Section C consists of 6 questions of 3 marks each
26. Number of students in each group subject to the given condition = HCF (60,84,108) ½
HCF (60,84,108) = 12 ½
60
Number of groups in Music = =5
12 ½
84
Number of groups in Dance = =7 ½
12
108 ½
Number of groups in Handicrafts = =9 ½
12
Total number of rooms required = 21
27. P(x) = 5x2 + 5x + 1 ½
−𝑏 −5
α+β = = = -1
𝑎 5 ½
𝑐 1
αβ = = ½
𝑎 5
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 - 2𝛼𝛽
1 ½
= (-1)2 – 2 ( )
5
2 3 ½
=1- =
5 5
1 1 ½
𝛼 +𝛽 = +
−1 −1
𝛼 𝛽
(𝛼+𝛽) (−1)
= = 1 = -5
𝛼𝛽
5
28. Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the original number = 10x + y
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s
Digit. ½
So the obtain by reversing the digits= 10y + x
According to the given condition.
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66 ½
i.e., x + y = 6 ---- (1)
We are also given that the digits differ by 2, ½
therefore, either x – y = 2 ---- (2) ½
or y – x = 2 ---- (3)
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. ½
In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. ½
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
1 1
Let √𝑥 be ‘m’ and √𝑦 be ‘n’, ½
Then the given equations become
2m + 3n = 2
½
4m - 9n = -1
3

( 2m + 3n = 2) X- 2 ⇒ −4𝑚 − 6𝑛 = −4 …(1)
4m - 9n = -1 4𝑚 − 9𝑛 = −1 …(2)
Adding (1) and (2)
1
We get −15𝑛 = −5 ⇒ 𝑛 = 3 ½

1
Substituting n = 3 in 2m + 3n = 2, we get ½
2m + 1 = 2
2m = 1
1
m= 2 1
1 1
m= 2 ⇒ √𝑥 = 2 ⇒ x = 4 and n = 3 ⇒ √𝑦 = 3 ⇒ y = 9
29.
∠OAB = 30°
∠OAP = 90° [Angle between the tangent and
the radius at the point of contact]
∠PAB = 90° - 30° =60° ½
AP = BP [Tangents to a circle from an external point]
∠PAB = ∠PBA [Angles opposite to equal sides of a triangle] ½
In ΔABP, ∠PAB + ∠PBA + ∠APB = 180° [Angle Sum Property]
60° + 60° + ∠APB = 180°
∠APB = 60° ½
∴ ΔABP is an equilateral triangle, where AP = BP = AB.
PA = 6 cm ½
In Right ΔOAP, ∠OPA = 30°
𝑂𝐴
tan 30° = 𝑃𝐴
1 𝑂𝐴 ½
=
√3 6
6
OA = = 2√3𝑐𝑚 ½
√3

Let ∠ TPQ = 𝜃
∠ TPO = 90° [Angle between the tangent and
the radius at the point of contact] ½
∠ OPQ = 90° - 𝜃
TP = TQ [Tangents to a circle from an external
point]
½
∠ TPQ = ∠ TQP = 𝜃 [Angles opposite to equal sides of a triangle] ½
In ΔPQT, ∠PQT + ∠QPT + ∠PTQ = 180° [Angle Sum Property] ½
𝜃 + 𝜃 + ∠PTQ = 180°
∠PTQ = 180° - 2 𝜃 ½
∠PTQ = 2 (90° - 𝜃) ½
∠PTQ = 2 ∠ OPQ [using (1)]
30. Given, 1 + sin2θ = 3 sin θ cos θ
Dividing both sides by cos2θ,
1
+ tan2θ = 3 tan 𝜃
cos2 𝜃
½
sec2θ + tan2θ = 3 tan 𝜃
½
1 + tan2θ + tan2θ = 3 tan 𝜃
½
1 + 2 tan2θ = 3 tan 𝜃
½
2 tan θ - 3 tan 𝜃 +1 = 0
2

If tan θ = x, then the equation becomes 2x2 -3x + 1 = 0
4

1
⇒ (𝑥 − 1)(2𝑥 − 1) = 0 x = 1 or 2
1
tan θ = 1 or 1
2
31.
Length Number of
CI Mid x d fd
[in mm] leaves (f)
118 – 126 3 117.5- 126.5 122 -27 -81

127 – 135 5 126.5– 135.5 131 -18 -90

136 – 144 9 135.5– 144.5 140 -9 -81

145 – 153 12 144.5 – 153.5 a = 149 0 0
153.5 – 162.5 158 9 45 2
154 – 162 5
163 – 171 4 162.5 – 171.5 167 18 72 ½
½
172 – 180 2 171.5 – 180.5 176 27 54
∑ 𝑓𝑑 −81
Mean = a + = 149 +
∑𝑓 40
= 149 – 2.025 = 146.975
Average length of the leaves = 146.975
SECTION D

Section D consists of 4 questions of 5 marks each

32. Let the speed of the stream be x km/h.
The speed of the boat upstream = (18 – x) km/h and
the speed of the boat downstream = (18 + x) km/h. 1
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 24
The time taken to go upstream = = hours
𝑠𝑝𝑒𝑒𝑑 18−𝑥
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 24
the time taken to go downstream = = hours 1
𝑠𝑝𝑒𝑒𝑑 18+𝑥
According to the question,
24 24
- =1 1
18−𝑥 18+𝑥

24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
x2 + 48x – 324 = 0
x = 6 or – 54 1
Since x is the speed of the stream, it cannot be negative.
Therefore, x = 6 gives the speed of the stream = 6 km/h.
1
Let the time taken by the smaller pipe to fill the tank = x hr.
Time taken by the larger pipe = (x – 10) hr ½
1
Part of the tank filled by smaller pipe in 1 hour =
𝑥
1 1
Part of the tank filled by larger pipe in 1 hour =
𝑥−10
3 75 ½
The tank can be filled in 9 8 = hours by both the pipes together.
8
8 ½
Part of the tank filled by both the pipes in 1 hour =
75
5

1 1 8
Therefore, + = ½
𝑥 𝑥−10 75
8x2 – 230x +750 = 0
30 1
x = 25,
8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by ½
the larger pipe will become negative, which is logically not possible.
Therefore, the time taken individually by the smaller pipe and ½
the larger pipe will be 25 and 25 – 10 =15 hours, respectively.

33. (a) Statement – ½
Given and To Prove – ½
Figure and Construction ½ 3
Proof – 1 ½
G
[b] Draw DG ‖ BE
𝐴𝐵 𝐴𝐸 ½
In Δ ABE, = [BPT]
𝐵𝐷 𝐺𝐸

CF = FD [F is the midpoint of DC] ---(i) ½
𝐷𝐹 𝐺𝐸
In Δ CDG, = = 1 [Mid point theorem]
𝐶𝐹 𝐶𝐸 ½
GE = CE ---(ii)
∠CEF = ∠CFE [Given]
CF = CE [Sides opposite to equal angles] ---(iii) ½
From (ii) & (iii) CF = GE ---(iv)
From (i) & (iv) GE = FD
𝐴𝐵 𝐴𝐸 𝐴𝐵 𝐴𝐸
∴ = ⇒ =
𝐵𝐷 𝐺𝐸 𝐵𝐷 𝐹𝐷
34.
Length of the pond, l= 50m, width of the pond, b = 44m
21
Water level is to rise by, h = 21 cm = m
100
21
Volume of water in the pond = lbh = 50 x 44 x m3 =462 m3 1
100
Diameter of the pipe = 14 cm
7
Radius of the pipe, r = 7cm = m
100
Area of cross-section of pipe = πr²
22 7 7 154 1
= x x = m2 ½
7 100 100 10000
Rate at which the water is flowing through the pipe, h = 15km/h = 15000 m/h
Volume of water flowing in 1 hour = Area of cross-section of pipe x height of water ½
coming out of pipe
154
= (10000 × 15000) 𝑚3 1
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑛𝑑
Time required to fill the pond = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑛 1 ℎ𝑜𝑢𝑟 1
462 ×10000
= = 2 hours
154 × 15000
Speed of water if the rise in water level is to be attained in 1 hour = 30km/h

6

Radius of the cylindrical tent (r) = 14 m
Total height of the tent = 13.5 m
Height of the cylinder = 3 m
Height of the Conical part = 10.5 m ½
Slant height of the cone (𝑙) = √ℎ2 + 𝑟 2
= √(10.5)2 + (14)2
= √110.25 + 196
1
= √306.25 = 17.5 m
Curved surface area of cylindrical portion
= 2πrh
22
= 2x ×14×3 1
7
= 264 m2
Curved surface area of conical portion
=πrl
22
= ×14×17.5 1
7
=770 m2 ½
Total curved surface area = 264 m2 + 770 m2 = 1034 m2
Provision for stitching and wastage = 26 m2
½
Area of canvas to be purchased = 1060 m2
Cost of canvas = Rate × Surface area ½
= 500 x 1060 = ₹ 5,30,000/-
35.
Number of Cumulative
Marks obtained
students frequency
20 – 30 p p
30 – 40 15 p + 15
40 – 50 25 p + 40 1
50 – 60 20 p + 60
60 – 70 q p + q + 60
70 – 80 8 p + q + 68 ½
80 - 90 10 p + q + 78 ½
90
p + q + 78 = 90
p + q = 12
𝑛
−𝑐𝑓
Median =(𝑙) +
2
.h
𝑓
45−(𝑝+40) ½
50 = 50 + . 10
20
45−(𝑝+40) ½
. 10 = 0
20
45 − (𝑝 + 40) = 0
½
P=5
½
5 + q = 12
q=7
𝑓1−𝑓0 1
Mode = 𝑙 + .h
2𝑓1−𝑓0−𝑓2

7

25−15
= 40 + . 10
2(25)−15−20
100
= 40 + = 40 + 6.67 = 46.67
15
SECTION E
36. (i) Number of throws during camp. a = 40; d = 12 1
𝑡11 = 𝑎 + 10𝑑
= 40 + 10 × 12
= 160 𝑡ℎ𝑟𝑜𝑤𝑠
(ii) a = 7.56 m; d = 9cm = 0.09 m ½
n = 6 weeks ½
tn = a + (n-1) d ½
= 7.56 + 6(0.09)
= 7.56 + 0.54 ½
Sanjitha’s throw distance at the end of 6 weeks = 8.1 m
(or)
a = 7.56 m; d = 9cm = 0.09 m ½
tn =11.16 m ½
tn = a + (n-1) d
11.16 = 7.56 + (n-1) (0.09) ½
3.6 = (n-1) (0.09)
3.6
n-1 = = 40
0.09
n = 41 ½
Sanjitha’s will be able to throw 11.16 m in 41 weeks.
(iii) a = 40; d = 12; n = 15
𝑛 ½
Sn = [2a + (n-1) d]
2
15
Sn = [2(40) + (15-1) (12)]
2
15
= [80 + 168]
2
15 ½
= [248] =1860 throws
2
37. (i) Let D be (a,b), then
Mid point of AC = Midpoint of BD
1+6 2+6 4+𝑎 3+𝑏 ½
( , )=( , )
2 2 2 2
4+a=7 3+ b = 8
a=3 b=5
Central midfielder is at (3,5) ½

8

(ii) GH = √(−3 − 3)2 + (5 − 1)2 = √36 + 16 = √52 = 2√13 ½
½
GK = √(0 + 3)2 + (3 − 5)2 = √9 + 4 = √13 ½
HK = √(3 − 0)2 + (1 − 3)2 = √9 + 4 = √13 ½
GK +HK = GH ⇒G,H & K lie on a same straight line
CJ = √(0 − 5) + (1 + 3) = √25 + 16 = √41
2 2 ½
½
CI =√(0 + 4)2 + (1 − 6)2 = √16 + 25 = √41
Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1)
5−4 −3+6 1 3 ½
Mid-point of IJ = ( , )=( , ) ½
2 2 2 2
C is NOT the mid-point of IJ

(iii) A,B and E lie on the same straight line and B is equidistant from A and E
⇒ B is the mid-point of AE
1+𝑎 4+𝑏 ½
( , ) = (2, −3) ½
2 2
1 + 𝑎 = 4 ; a = 3. 4+b = -6; b = -10 E is (3,-10)
38. 80
(i) tan 45° = 𝐶𝐵 ⇒ CB = 80m 1
80 ½
(ii) tan 30° =
𝐶𝐸 ½
1 80 ½
⇒ = ½
√3 𝐶𝐸

⇒ CE = 80√3
Distance the bird flew = AD = BE = CE-CB = 80√3 – 80 = 80(√3 -1) m
½
(or) ½
80
tan 60° =
𝐶𝐺
80 ½
⇒ √3 = ½
𝐶𝐺

80
⇒ CG =
√3
Distance the ball travelled after hitting the tree =FA=GB = CB -CG

80 1
GB = 80 - = 80 (1 - )m
√3 √3
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 20(√3 + 1) ½
(iii) Speed of the bird = = m/sec
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 2
20(√3 + 1) ½
= x 60 m/min = 600(√3 + 1) m/min
2

9